# 7.3: Exercises

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## Exercise $$\PageIndex{1}$$

Use the horizontal line test to determine if the function is one-to-one.

1. $$f(x)=x^2+2x+5$$
2. $$f(x)=x^2-14 x+29$$
3. $$f(x)=x^3-5x^2$$
4. $$f(x)=\dfrac{x^2}{x^2-3}$$
5. $$f(x)=\sqrt{x+2}$$
6. $$f(x)=\sqrt{|x+2|}$$
1. no (that is: the function is not one-to-one)
2. yes
3. no
4. no
5. no
6. no
7. yes
8. no

## Exercise $$\PageIndex{2}$$

Find the inverse of the function $$f$$ and check your solution.

1. $$f(x)=4x+9$$
2. $$f(x)=-8x-3$$
3. $$f(x)=\sqrt{x+8}$$
4. $$f(x)=\sqrt{3x+7}$$
5. $$f(x)=6\cdot \sqrt{-x-2}$$
6. $$f(x)=x^3$$
7. $$f(x)=(2x+5)^3$$
8. $$f(x)=2\cdot x^3+5$$
9. $$f(x)=\dfrac{1}{x}$$
10. $$f(x)=\dfrac{1}{x-1}$$
11. $$f(x)=\dfrac{1}{\sqrt{x-2}}$$
12. $$f(x)=\dfrac{-5}{4-x}$$
13. $$f(x)=\dfrac{x}{x+2}$$
14. $$f(x)=\dfrac{3x}{x-6}$$
15. $$f(x)=\dfrac{x+2}{x+3}$$
16. $$f(x)=\dfrac{7-x}{x-5}$$
17. $$f$$ given by the table: $$\begin{array}{|c||c|c|c|c|c|c|} \hline x & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline \hline f(x) & 3 & 7 & 1 & 8 & 5 & 2 \\ \hline \end{array}$$
1. $$f^{-1}(x)=\dfrac{x-9}{4}$$
2. $$f^{-1}(x)=-\dfrac{x+3}{8}$$
3. $$f^{-1}(x)=x^{2}-8$$
4. $$f^{-1}(x)=\dfrac{x^{2}-7}{3}$$
5. $$f^{-1}(x)=-\left(\dfrac{x}{6}\right)^{2}-2=\dfrac{-x^{2}-72}{36}$$
6. $$f^{-1}(x)=\sqrt[3]{x}$$
7. $$f^{-1}(x)=\dfrac{\sqrt[3]{x}-5}{2}$$
8. $$f^{-1}(x)=\sqrt[3]{\dfrac{x-5}{2}}$$
9. $$f^{-1}(x)=\dfrac{1}{x}+1=\dfrac{1+x}{x}$$
10. $$f^{-1}(x)=\left(\dfrac{1}{x}\right)^{2}+2=\dfrac{1+2 x^{2}}{x^{2}}$$
11. $$f^{-1}(x)=\dfrac{5}{y}+4=\dfrac{5+4 y}{y}$$
12. $$f^{-1}(x)=\dfrac{5}{y}+4=\dfrac{5+4 y}{y}$$
13. $$f^{-1}(x)=\dfrac{2 x}{1-x}$$
14. $$f^{-1}(x)=\dfrac{6 x}{x-3}$$
15. $$f^{-1}(x)=\dfrac{2-3 x}{x-1}$$
16. $$f^{-1}(x)=\dfrac{5 x+7}{x+1}$$
17. $$\begin{array}{|c||c|c|c|c|c|c|} \hline x & 3 & 7 & 1 & 8 & 5 & 2 \\ \hline \hline f^{-1}(x) & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}$$

## Exercise $$\PageIndex{3}$$

Restrict the domain of the function $$f$$ in such a way that $$f$$ becomes a one-to-one function. Find the inverse of $$f$$ with the restricted domain.

1. $$f(x)=x^2$$
2. $$f(x)=(x+5)^2+1$$
3. $$f(x)=|x|$$
4. $$f(x)=|x-4|-2$$
5. $$f(x)=\dfrac{1}{x^2}$$
6. $$f(x)=\dfrac{-3}{(x+7)^2}$$
7. $$f(x)=x^4$$
8. $$f(x)=\dfrac{(x-3)^4}{10}$$
1. restricting to the domain $$D=[0, \infty)$$ gives the inverse $$f^{-1}(x)=\sqrt{x}$$
2. restricting to the domain $$D=[-5, \infty)$$ gives the inverse $$f^{-1}(x)=\sqrt{x-1}-5$$
3. restricting to the domain $$D=[0, \infty)$$ gives the inverse $$f^{-1}(x)=x$$
4. restricting to the domain $$D=[4, \infty)$$ gives the inverse $$f^{-1}(x)=x+6$$
5. restricting to the domain $$D=[0, \infty)$$ gives the inverse $$f^{-1}(x)=\sqrt{\dfrac{1}{x}}$$
6. restricting to the domain $$D=[-7, \infty)$$ gives the inverse $$f^{-1}(x)=\sqrt{-\dfrac{3}{x}}-7$$
7. restricting to the domain $$D=[0, \infty)$$ gives the inverse $$f^{-1}(x)=\sqrt[4]{x}$$
8. restricting to the domain $$D=[3, \infty)$$ gives the inverse $$f^{-1}(x)=3+\sqrt[4]{10 x}$$

## Exercise $$\PageIndex{4}$$

Determine whether the following functions $$f$$ and $$g$$ are inverse to each other.

1. $$f(x)=x+3, \quad g(x)=x-3$$,
2. $$f(x)=-x-4, \quad g(x)=4-x$$,
3. $$f(x)=2x+3, \quad g(x)=x-\dfrac{3}{2}$$,
4. $$f(x)=6x-1, \quad g(x)=\dfrac{x+1}{6}$$,
5. $$f(x)=x^3-5, \quad g(x)= 5+\sqrt[3]{x}$$,
6. $$f(x)=\dfrac{1}{x-2}, \quad g(x)=\dfrac{1}{x}+2$$.
1. yes (that is: the functions f and g are inverses of each other)
2. no
3. no
4. yes
5. no
6. yes

## Exercise $$\PageIndex{5}$$

Draw the graph of the inverse of the function given below.

1. $$f(x)=\sqrt{x}$$
2. $$f(x)=x^3-4$$
3. $$f(x)=2x-4$$
4. $$f(x)=2^x$$
5. $$f(x)=\dfrac{1}{x-2}$$ for $$x>2$$
6. $$f(x)=\dfrac{1}{x-2}$$ for $$x<2$$